Thermodynamics with styrofoam cooler

[physikx]

Homework Statement

A cubical styrofoam cooler 80cm on a side and 2.0 cm thick contains 2.0kg of ice at 0C. If it takes four hours for the ice to melt what is the outside temperature? K(st)=.02 w/m*K

Homework Equations

H=-KA (ΔT/ΔX)
(conductive heat flow)

The Attempt at a Solution

Area: 6(a)^2= 6(.8)^2= 3.8m^2
Δx=.02m
K(st)=.02 w/m*K

H=-.02 w/m*K(3.8m^2) (ΔT/.02m)So I used the formula for the conductive heat flow, but I still have two unknowns. I know H is ΔQ/Δt, but I am asked for the time. I had two questions in regards to this problem:
1. for the area do I find the total surface area or just one side?
2. what temperature change ΔT should I set the ΔQ formula if I am assuming the ice is melting?

Thanks for the help!

 

[ehild]

So I used the formula for the conductive heat flow, but I still have two unknowns. I know H is ΔQ/Δt, but I am asked for the time. I had two questions in regards to this problem:
1. for the area do I find the total surface area or just one side?
2. what temperature change ΔT should I set the ΔQ formula if I am assuming the ice is melting?

Thanks for the help!

First of all, what do the symbols in the formula for the conductive heat flow mean?
Can you derive that formula?


ehild

 

[physikx]

ehild, thank you for your reply and I apologize for this far too late of response. I figured out the answer to this problem after grinding the problem for a while. I am done with my final for the first part of physics! I can not wait for the second part in a week :) Thanks again and talk to you soon.

physikx

 

[ehild]

Congratulation!

ehild

 

[asteriatic]

I would approach this problem by first clarifying some assumptions and variables. The problem states that the cooler is 80 cm on a side and 2.0 cm thick, but does not specify the material of the cooler. For this solution, I will assume that the cooler is made of styrofoam, as stated in the problem title. Additionally, the problem does not specify the initial temperature of the outside environment, so I will use "T" to represent the unknown outside temperature.

Now, to answer your questions:

1. For the area, you should use the total surface area of the cooler, which is 6(a)^2, as you have correctly calculated. This is because heat can flow through all six sides of the cooler.

2. For the temperature change, you can use the ΔT of 0°C to represent the ice melting. This is because the ice will remain at 0°C until it has completely melted.

Now, we can use the formula for conductive heat flow, H=-KA (ΔT/ΔX), to solve for the outside temperature (T):

ΔQ/Δt = -KA (ΔT/ΔX)
(2.0kg * 334 kJ/kg) / (4 hours * 3600 seconds/hour) = (-.02 W/m*K) * (3.8 m^2) * (0°C - T) / (0.02 m)
0.185 kJ/s = -0.76 (0°C - T)
T = 0.24°C

Therefore, the outside temperature is approximately 0.24°C. This means that the outside environment must be slightly above freezing for the ice to melt in four hours.